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Section 9.2: Vectors
Practice HW from Stewart Textbook (not to hand in)
p. 649 # 7-20
Vectors in 2D and 3D Space
Scalars are real numbers used to denote the amount (magnitude) of a quantity. Examples
include temperature, time, and area.
Vectors are used to indicate both magnitude and direction. The force put on an object or
the velocity a pitcher throws a baseball are examples.
Notation for Vectors
Suppose we draw a directed line segment between the points P (called the initial point)
and the point Q (called the terminal point).
Q
P

We denote the vector between the points P and Q as v  PQ . We denote the length or
magnitude of this vector as

Length of v = v  PQ
2
We would like a way of measuring the magnitude and direction of a vector. To do this,
we will example vectors both in the 2D and 3D coordinate planes.
Vectors in 2D Space
Consider the x-y coordinate plane. In 2D, suppose we are given a vector v with initial
point at the origin (0, 0) and terminal point given by the ordered pair (v1 , v2 ) .
v2
v1
The vector v with initial point at the origin (0, 0) is said to be in standard position. The
component for of v is given by v =  v1 ,v2  .
Example 1: Write in component form and sketch the vector in standard position with
terminal point (1, 2).
Solution:
█
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Vectors in 3D Space
Vectors is 3D space are represented by ordered triples v =  v1 , v2 , v3  . A vector v is
standard position has its initial point at the origin (0,0,0) with terminal point given by the
ordered triple (v1 , v2 , v3 ) .
z
y
x
Example 2: Write in component form and sketch the vector in standard position with
terminal point (-3, 4, 2).
Solution:
█
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Some Facts about Vectors
1. The zero vector is given by 0 = < 0, 0 > in 2D and 0 = < 0, 0, 0 > in 3D.
2. Given the points A  ( x1 , y1 ) and B  ( x2 , y2 ) in 2D not at the origin.
y
B  ( x2 , y 2 )

a  AB =  a1 , a2  x2  x1 , y2  y1 
A  ( x1 , y1 )
x

The component for the vector a is given by a  AB =  a1 , a2  x2  x1 , y2  y1  .
Given the points A  ( x1 , y1 , z1 ) and B  ( x2 , y2 , z 2 ) in 3D not at the origin.
B  ( x2 , y 2 , z 2 )
z

A  ( x1 , y1 , z1 )
a  AB =  a1 , a2 , a3  x2  x1 , y 2  y1 , z 2  z1 
y
x

The component for the vector a is given by a  AB =  a1 , a2 , a3  x2  x1 , y 2  y1 , z 2  z1  .
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3. The length (magnitude) of the 2D vector a =  a1 , a2  is given by
|a| =
a12  a22
The length (magnitude) of the 3D vector a =  a1 , a2 , a3  is given by
|a| =
a12  a22  a32
4. If | a | = 1, then the vector a is called a unit vector.
5. | a | = 0 if and only if a = 0.
Example 3: Given the points A(3, -5) and B(4,7).

a. Find a vector a with representation given by the directed line segment AB .
b. Find the length | a | of the vector a.

c. Draw AB and the equivalent representation starting at the origin.
Solution:
█
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Example 4: Given the points A(2, -1, -2)and B(-4, 3, 7)..

a. Find a vector a with representation given by the directed line segment AB .
b. Find the length | a | of the vector a.

c. Draw AB and the equivalent representation starting at the origin.
Solution:
█
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Facts and Operations With Vectors 2D
Given the vectors a =  a1 , a2  and b =  b1 ,b2  , k be a scalar. Then the following
operations hold.
1. a + b =  a1 , a2    b1 , b2  a1  b1 , a2  b2  . (Vector Addition)
a - b =  a1 , a2    b1 , b2  a1  b1 , a2  b2  . (Vector Subtraction)
2. k a = k  a1 , a2  ka1 , ka2  . (Scalar-Vector Multiplication)
3. Two vectors are equal if and only if their components are equal, that is, a = b if and
only if a1  b1 and a2  b2 .
Facts and Operations With Vectors 3D
Given the vectors a =  a1 , a2 , a3  and b =  b1 , b2 , b3  , k be a scalar. Then the
following operations hold.
1. a + b =  a1 , a 2 , a3    b1 , b2 , b3  a1  b1 , a 2  b2 , a3  b3  . (Vector Addition)
a - b =  a1 , a 2 , a3    b1 , b2 , b3  a1  b1 , a 2  b2 , a3  b3  . (Vector Subtraction)
2. k a = k  a1 , a2 , a3  ka1 , ka2 , ka3  . (Scalar-Vector Multiplication)
3. Two vectors are equal if and only if their components are equal, that is, a = b if and
only if a1  b1 , a2  b2 , and a3  b3 .
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Example 5: Given the vectors a =  3,1  and b =  1,2  , find
a. a + b
c. 3 a – 2 b
b. 2 b
d. | 3 a – 2 b |
Solution:
█
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Unit Vector in the Same Direction of the Vector a
Given a non-zero vector a, a unit vector u (vector of length one) in the same direction as
1
the vector a can be constructed by multiplying a by the scalar quantity
, that is,
|a|
forming
u
Multiplying the vector | a | by
1
a
a
|a|
|a|
1
to get the unit vector u is called normalization.
|a|
y
a
u
x
10
Example 6: Given the vector a = < -4, 5, 3>.
a. Find a unit vector in the same direction as a and verify that the result is indeed a unit vector.
b. Find a vector that has the same direction as a but has length 10.
Solution: Part a) To compute the unit vector u in the same direction of a = < -4, 5, 3>, we
first need to find the length of a which is given by
|a|=
(4) 2  (5) 2  (3) 2  16  25  9  50  25  2  5 2
Then
u
1
1
4
5
3
4
1
3
a
 4, 5, 3  ,
,
 ,
,
.
|a|
5 2
5 2 5 2 5 2
5 2 2 5 2
For u to be a unit vector, we must show that | u | = 1. Computing the length of | u | we obtain
2
2
2
4 

 1 
 3 
| u |  
 
 
 
 5 2
 2
5 2 
16
1
9
16 25 9
50
 




 1 1
25  2 2 25  2
50 50 50
50
Part b) Since the unit vector u found in part a has length 1 is in the same direction of a,
multiplying the unit vector u by 10 will give a vector, which we will call b, with a length of
10, in the same direction of a. Thus,
b  10u  10  
4
5 2
,
1
,
3
2 5 2
 
40
5 2
,
10
,
30
2 5 2
 
8
2
,
10
2
,
6
2

The following graph shows the 3 vectors on the same graph, where you can indeed see they
are all pointing in the same direction (the unit vector u is in red, the given vector a in blue,
and the vector b in green.
█
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Standard Unit Vectors
In 2D, the unit vectors < 1, 0 > and < 0, 1 > are the standard unit vectors. We denote these
vectors as i = < 1, 0 > and j = < 0, 1 >. The following represents their graph in the x-y
plane.
y
(0, 1)
j
i
(1, 0)
x
Any vector in component form can be written as a linear combination of the standard unit
vectors i and j. That is, the vector a =  a1 , a2  in component for can be written
a =  a1 , a2  a1 ,0    0, a2  a1  1,0  a2  0,1  = a1 i + a 2 j
in standard unit vector form. For example, the vector < 2, -4 > in component form can be
written as
 2,4  2i  4 j
in standard unit vector form.
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In 3D, the standard unit vectors are i = < 1, 0, 0 > , j = < 0, 1, 0 >, and k = < 0, 0, 1 >.
z
(0, 0, 1)
k
j
(1, 0, 0)
i
(0, 1, 0)
y
x
Any vector in component form can be written as a linear combination of the standard unit
vectors i and j and k. That is, That is, the vector a =  a1 , a2 , a3  in component for can be
written
a=
 a1 , a2 , a3  a1 ,0,0    0, a2 ,0    0,0, a3  a1  1,0,0  a2  0,1,0  a3  0,0,1 
= a1 i + a 2 j + a 3 k
in standard unit vector from. For example the vector the vector < 2, -4, 5 > in component
form can be written as
 2,4, 5  2i  4 j  5k
in standard unit vector form.
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Example 7: Given the vectors a  i  2 j  3k , b  2i  2 j  k , and c  4i  4k , find
a. a – b
b. | a – b |
1
c. | 5a  3b  c |
2
Solution: